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I recently read the answer to a question regarding the difference between a tree and a spanning tree. The following is the link: Difference between a tree and spanning tree?!

Now we know that the total possible trees for a graph = n^(n-2). Therefore, for a graph containing 4 nodes, number of trees possible = 4^(4-2) = 16

Graph and all of its possible trees

Looking at the picture, it is clear that all of the 16 trees contain all 4 nodes present in the original graph, i.e. there is not a single tree with less than 4 nodes. But the answer provided in the question to the link shared above says otherwise. I am pretty confused by now. Any response would be greatly appreciated.

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  • $\begingroup$ A tree is connected. The formula for counting trees on 4 vertices (nodes) counts exactly that: the trees that contain exactly 4 nodes. A subgraph does not have to contain all nodes of the original graph, so can still possibly be a tree without containing all the vertices of the original. $\endgroup$ – Morgan Rodgers Jan 21 at 19:02
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    $\begingroup$ Cayley's formula says that there are $n^{n-2}$ labeled trees on $n$ vertices, i.e. each of those trees must contain all $n$ vertices, hence they would span the complete graph. I think you have a misunderstanding of this formula, especially because I'm not sure what you mean by the "total possible trees for a graph". If you mean subgraphs which are trees, then Cayley's formula does not count that. $\endgroup$ – Kevin Long Jan 21 at 19:32
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In order to close the topic :

Cayley's formula counts the number of labelled trees on $n$ vertices, hence of spanning tree of $K_n$ ( the complete graph on $n$ vertices).

Cayley's formula does not count all possibles trees, the spanning tress of subgraphs of $K_n$. If you want to do count this, then you need to iterate through all subgraphs of $K_n$ : Counting the number of labelled spanning trees of each $K_{n-k}$, multiplied by the number of labelled $K_{n-k}$:

$$ T = \sum_{k=0}^{n-1} \binom{n}{k} (n-k)^{n-k-2} $$ I don't know if this sum can be simplified.

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